BRUFACE Vibrations and Acoustics MA1 Academic year 17-18 Cédric Dumoulin (cedumoul@ulb.ac.be) Arnaud Deraemaeker (aderaema@ulb.ac.be) Exercise 1 Session 1 : Fundamental concepts Consider the following one-degree-of-freedom (1 DOF) system Write the equation of motion in the time domain. mẍ + cẋ + kx = f Give the expression of ω n and of ξ ω n = k/m ξ = c mk For this system a) Give the expression of the impulse response and represent it using the following numerical values: m = 1 kg, k = 16 N/m, c =.1 Ns/m) with ω d = ω n 1 ξ h(t) = 1 mω d e ξωnt sin ω d t t=linspace(,1,1); k=16; m=1; c=.1; wn=sqrt(k/m); xi=c/(*sqrt(k*m)); wd=wn*sqrt(1-xiˆ); h=1/(m*wd)*exp(-xi*wn*t).*sin(wd*t); figure; set(gca, Fontsize,15); plot(t,h); xlabel( time ); ylabel( Impulse response h(t) )
b) Give the expression of the harmonic forced response and represent it using the Bode diagram H(ω) = X F = 1 m (ω n ω + jξωω n ) w=linspace(,,1); H=1./(m*(wnˆ-w.ˆ+*j*xi*w*wn)); figure; set(gca, Fontsize,15); subplot(,1,1); plot(w,*log1(abs(h))); xlabel( \omega );ylabel( Mag(dB) ) subplot(,1,); plot(w,angle(h)*18/pi); xlabel( \omega );ylabel( Phase(deg) ) c) Plot the Nyquist diagram and indicate the resonant frequency as well as some intermediary frequencies on the diagram. Comment. RE=real(H); IM=imag(H); figure; set(gca, Fontsize,15); plot(re(:),im(:)); xlabel( Re(H) ); ylabel( Im(H) ) ind=[197 198 199 1 3]; for i=ind text(re(i),im(i),[ \omega= numstr(w(i))]) The Nyquist plot looks like a circle, we clearly see that most values on the circle correspond to a frequency range in the close vicinity of the resonant frequency of the system. d) Repeats points a),b) and c) with the following successive values of damping: c =.1 Ns/m, c =.5 Ns/m, c = 1 Ns/m. What are the corresponding values of ξ? Plot the respective responses on the same plot t=linspace(,1,1); k=16; m=1; c=.1; wn=sqrt(k/m); ci=[.1.5 1]; color= rgb ; figure; set(gca, Fontsize,15) for i=1:length(ci) c=ci(i); xi=c/(*sqrt(k*m)); wd=wn*sqrt(1-xiˆ); h=1/(m*wd)*exp(-xi*wn*t).*sin(wd*t); plot(t,h,color(i)); hold on; figure; set(gca, Fontsize,15) for i=1:length(ci) c=ci(i); xi=c/(*sqrt(k*m)); wd=wn*sqrt(1-xiˆ); H=1./(m*(wnˆ-w.ˆ+*j*xi*w*wn)); subplot(,1,1); plot(w,*log1(abs(h)),color(i)); hold on xlabel( \omega );ylabel( Mag(dB) )
subplot(,1,); plot(w,angle(h)*18/pi,color(i)); hold on xlabel( \omega );ylabel( Phase(deg) ) figure; set(gca, Fontsize,15) for i=1:length(ci) c=ci(i); xi=c/(*sqrt(k*m)); wd=wn*sqrt(1-xiˆ); H=1./(m*(wnˆ-w.ˆ+*j*xi*w*wn)); RE=real(H); IM=imag(H); plot(re(:),im(:),color(i)); hold on xlabel( Re(H) ); ylabel( Im(H) ) c =.1 ξ =.15 c =.5 ξ =.65 c = 1 ξ = 1.5 e) Write the relationship between the impulse response and the forced harmonic response. Verify numerically that it is valid. H(ω) = h(τ)e jωτ dτ t=linspace(,15,1); dt=t()-t(1); k=16; m=1; c=.1; wn=sqrt(k/m); xi=c/(*sqrt(k*m)); wd=wn*sqrt(1-xiˆ); h=1/(m*wd)*exp(-xi*wn*t).*sin(wd*t); a=fft(h)*dt; % see note below w=linspace(,1/dt**pi,length(a)); H=1./(m*(wnˆ-w.ˆ+*j*xi*w*wn)); figure; plot(w,*log1(abs(a))); hold on; plot(w,*log1(abs(h(:))), g ); set(gca, Xim,[ ]) leg( FT of h(t), H ); Note : the discrete Fourier transform (DFT) is defined as : c n = 1 T T u(t) e inω t dt When the signal u(t) is sampled at m regular time intervals with time spacing t, for a small value of t we have m m c n T t u(j t) e inωj t = t u(j t) e inωj t = ( t)fft(u) j= where fft(u) is the discrete Fourier transform of a sampled signal as computed in Matlab using the Fast Fourier Transform algorithm (FFT). For the limit of T, c n T is the continuous Fourier transform. Therefore a good approximation of the continuous Fourier transform can be obtained by taking a long period T, a short time spacing t, and performing the discrete fourier transform f f t(u) in Matlab, then multiplying by t. j=
Exercise Consider a bridge crane consisting of a simply supported beam to which the lifting system is attached, as shown in Figure 1(a). The system can be modeled as a simply supported beam to which a mass m (equal to the total mass of the lifting system and the weight) is attached, as shown in Figure (a). (a) Bridge crane (b) 1DOF model. Figure 1: Bridge crane and equivalent SDOF system EI;úS m (a) Model of the bridge crane EI;úS m W (b) Model of the bridge crane with an additional mass W Figure : Bridge crane model with and without additional mass The beam is a steel IPE 75x196 (see tables) of length = 1m (E = 1GP a, ρ = 78kg/m 3 ), and the mass m is 5 Tons. Assuming that the mass of the beam is small with respect to the mass m, which is located at position x = /: 1. Determine the stiffness k, and the mass M of the equivalent 1 DOF system represented in Figure 1(b) (use the tables provided). Use the energy method in order to determine
the part of the equivalent mass due to the mass of the beam. Give the expression of the eigenfrequency of the equivalent 1 DOF system. For the stiffness, from the tables, we find : k = 48EI 3 = 48 1 19.4 1 3 1 3 = 4kN/m For the equivalent mass, the kinetic energy is computed on a half beam. The static displacement profile under a load P is given by y(x) = P ( ) 3 x 1EI 4 x3 < x < / and the velocity profile is therefore assumed to be ( ) v(x) = 3 x 4 4 x3 x 3 1 so that the velocity of the center point where the mass m is attached is v(/) = x 1 The kinetic energy of the half beam is given by [ E k = 1 which requires to compute the following integral = [ 3 4 x 3 = 16 ρ 16 ( ) 3 x dx] 6 4 x3 ( ) 3 x 4 x3 dx ( 9 4 x + x 6 3 ) 16 x 4 dx + x7 7 3 ] 1 x 5 = 37 18 + 7 896 37 3 = 177 11 m a1 = 16ρ 6 17 7 11 = 17 7 ρ m a = 17 17 ρ = 35 35 m beam The additional mass due to a half-beam is therefore m a1 = 16ρ 6 17 7 11 = 17 7 ρ x 1
And for the full beam, taking into account the numerical values, we have m a = 17 17 ρ = 35 35 m beam = 17 35 78 51 1 4 1 = 951kg The natural frequency of the system is therefore: f n = 1.4 1 7 π 5 + 951 = 1.15 (Hz) Hint for the equivalent mass due to the mass of the beam: Assume that the displacement profile of the beam is proportional to the static displacement of the beam due to a point force in the middle of the beam compute the associated kinetic energy and identify the equivalent mass by comparing with the kinetic energy of the 1 DOF system. Plot the shape of the frequency response function of the equivalent 1 DOF system assuming a damping value of ξ =.1. What is the amplification factor with respect to the static response? How many eigenfrequencies are present? How many eigenfrequencies would be present if you were to plot the frequency response function of the real system? k=.4*1e7; m = 5951 ; xi=.1; b=*xi*sqrt(k*m); w=linspace(,**pi,1); figure; plot(w/(*pi), *log1(abs(1./(k-w.ˆ*m+j*w*b)))) xlabel( Frequency (Hz) ); ylabel ( Disp center of beam (db) ) The static displacement under unit force is The amplification factor is X = 1/k = 4.13 1 8 m 1 ξ = 5 so that for a dynamic force of 1kN, the amplitude of the displacement at the center of the beam will be X =.64mm. For the real system, there is an infinite number of eigenfrequencies due to the fact that the beam is a continuous system. 3. Assume that an additional weight of mass W = 1kg is attached to the lifting system (Figure (b)). At time t =, The rope attaching this additional mass is cut. Plot the shape of the evolution of the displacement in the vertical direction at position x = / as a function of time (assume a damping value of ξ =.1 and use the equivalent 1 DOF representation). Indicate the value of the period of oscillation. How many oscillations are needed to reduce the amplitude of vibration by approximately one half? Use the logarithmic decrement method to determine the damping coefficient. If we consider that the system is at equilibrium before we add the mass W, we take the displacement of the center of the beam as the reference zero displacement. Adding the mass of 1 kg results in an initial displacement of: x = F/k = 1 9.81/.4 1 7 =.45mm
and an initial zero velocity. The general form of the solution is: ( x(t) = e ξωnt x cos ω d t + x ) + ω n ξx sin ω d t ω d and for a zero initial velocity : ( x(t) = e ξωnt x cos ω d t + ω ) nξx sin ω d t ω d t=linspace(,5,1); k=.4*1e7; m = 5951 ; xi=.1; b=*xi*sqrt(k*m); x=4.5e-4; wn=sqrt(k/m); wd=wn*sqrt(1-xiˆ); x=exp(-xi*wn*t).*(x*cos(wd*t)+wn/wd*xi*x*sin(wd*t)); figure; plot(t,x); xlabel( time(s) ); ylabel( Amplitude (m) ) The period of oscillation is given by T = 1 f = π ω d =.985s And using the logarithmic decrement method : x=x; n=; for i=1:n [ind(i)]=find(x==max(x)); x=x(ind(i):); j=find(x<); x=x(j:); max=x(ind) xi=1/(*pi*(n-1))*log(max(1)/max()); we can check that the value is very close to ξ =.1.